Operational amplifier

An operational amplifier is a differential electronic amplifier whose unloaded voltage is modeled in its linear regime by $v_o=A(s)(v_{+}-v_{-})$. Its large open-loop gain, large input impedance and small output impedance make negative feedback useful for setting a reproducible transfer function with external components. Supply voltages bound the available output swing; the linear model does not apply after saturation.

An op amp integrator places a resistor $R$ between the input source and the inverting input of an operational amplifier, and a feedback capacitor $C$ between that input and the output, with the non-inverting input grounded. For an ideal operational amplifier, Kirchhoff's current law gives $v_{\mathrm{in}}/R=-Cd{v}_o/dt$, henceThe initial capacitor charge supplies the integration constant. A parallel feedback resistor $R_f$ replaces the transfer by $H(s)=-(R_f/R)/(1+s{R}_{f}C)$: the direct-current gain becomes finite while integration is recovered when $|s{R}_{f}C|\gg 1$. This limits offset-driven saturation. ^[ref]

Consider an op amp integrator with infinite input impedance, an internal controlled voltage $u=-A(s)v$ where $v$ is the inverting-input voltage, and a series output impedance equal to a resistance $r_o$. Assume no external output load. Writing $y$ for the output, Kirchhoff's current law givesEliminating $u$ and $v$ yieldsThe expression at $r_o=0$ is obtained by its limit. In particular, constant gain $A_0$ and $r_o=0$ give $H(s)=-A_0/[1+sRC(1+A_0)]$: a finite direct-current gain followed by an integrating regime. For the dominant-pole model $A(s)=A_0/(1+s/{\omega }_a)$ and $r_o=0$,If $A_0\gg 1$ and $A_0{\omega }_{a}RC\gg 1$, there are separated turnover frequencies near $1/(A_{0}RC)$ and $A_0{\omega }_a$. The gain magnitude is respectively constant, proportional to ${\omega }^{-1}$ and proportional to ${\omega }^{-2}$. With the convention $e^{i\omega t}$ a continuous phase choice passes from $180^{\circ }$ through $90^{\circ }$ to $0^{\circ }$. If $r_o>0$ and $A(i\omega )\to 0$, capacitive feedthrough instead gives the ultimate limit $r_o/(R+r_o)$. Specifying finite gain alone therefore does not specify a unique high-frequency response.

The slew rate bounds the large-signal output-voltage change per unit time. It is distinct from a small-signal gain-bandwidth product. A sinusoidal output of amplitude $V_p$ and angular frequency $\omega$ needs $\omega V_p$ below the available slew rate to avoid this source of distortion.

An input bias current is a direct-current component drawn by an input terminal of a real operational amplifier. Through a source resistance it produces a voltage error, and through the feedback capacitor of an op amp integrator it can produce an output ramp. Its magnitude and sign depend on the amplifier technology and operating conditions.

The input offset voltage of an operational amplifier is the differential direct-current input voltage needed to make its output zero at a specified operating point. Its effect resembles a small unwanted input signal and can cause sustained output drift in an op amp integrator.

The open-loop gain of an operational amplifier is the ratio of its output voltage to its differential input voltage before an external feedback control connection is applied. A common small-signal model is $A(s)=A_0/(1+s/{\omega }_a)$, with a large direct-current gain $A_0$ and a dominant pole at $-{\omega }_a$. Above ${\omega }_a$ its magnitude falls approximately inversely with angular frequency.

For a dominant-pole voltage-feedback operational amplifier with $A(s)=A_0/(1+s/{\omega }_a)$ and $A_0\gg 1$, the unity-gain angular frequency is approximately ${\omega }_t=A_0{\omega }_a$. In the range where this pole dominates, a roughly constant closed-loop noise gain $G_N$ gives a bandwidth approximately $f_t/G_N$. This approximation depends on the amplifier model and feedback control circuit; it is not a universal statement about all amplifiers or all frequency-dependent feedback networks. ^[ref]

An ideal operational amplifier has infinite open-loop gain and input impedance, zero output impedance, unlimited response speed and zero input offset voltage and input bias current. Its input currents vanish. In a stable linear circuit with negative feedback and a finite output, its differential input voltage tends to zero as the open-loop gain tends to infinity. Thus $i_{+}=i_{-}=0$ and $v_{+}=v_{-}$ are the familiar conditional circuit rules. The equality of the input voltages cannot be imposed on an open-loop, unstable or saturated circuit.

Connect a resistor $R_{\mathrm{in}}$ from the input source to the inverting input, a feedback resistor $R_f$ from output to that same input, and ground the non-inverting input. For ideal linear operation the inverting node is a virtual ground. Kirchhoff's current law gives $V_i/R_{\mathrm{in}}+V_o/R_f=0$, yielding the displayed voltage gain. The source sees input impedance $R_{\mathrm{in}}$, even though the amplifier input itself has infinite input impedance.